The Stacks project

Proof. Assume (1) and let $U, V$ be as in (2). It follows from Lemma 37.60.1 that $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is pseudo-coherent. Hence it suffices to prove that the property of a ring map being "of finite tor dimension" satisfies conditions (1)(a), (b), (c) of Morphisms, Definition 29.14.1. These properties follow from More on Algebra, Lemmas 15.66.11, 15.66.14, and 15.66.16. Some details omitted. $\square$

Proof. This translates into the following algebra result: Let $A \to B$ be a perfect ring map. Let $A \to A'$ be flat. Then $A' \to B \otimes _ A A'$ is perfect. This result for pseudo-coherent ring maps we have seen in Lemma 37.60.3. The corresponding fact for finite tor dimension follows from More on Algebra, Lemma 15.66.14. $\square$

Proof. This translates into the following algebra result: If $A \to B \to C$ are composable perfect ring maps then $A \to C$ is perfect. We have seen this is the case for pseudo-coherent in Lemma 37.60.4 and its proof. By assumption there exist integers $n$, $m$ such that $B$ has tor dimension $\leq n$ over $A$ and $C$ has tor dimension $\leq m$ over $B$. Then for any $A$-module $M$ we have

and the spectral sequence of More on Algebra, Example 15.62.4 shows that $\text{Tor}^ A_ p(M, C) = 0$ for $p > n + m$ as desired. $\square$

Proof. The implication (2) $\Rightarrow $ (1) is More on Algebra, Lemma 15.82.4. The converse follows from the fact that a pseudo-coherent morphism is locally of finite presentation, see Lemma 37.60.5. $\square$

Proof. See More on Algebra, Lemma 15.82.5. $\square$

Proof. Since a regular immersion is a Koszul-regular immersion, see Divisors, Lemma 31.21.2, it suffices to prove the second statement. This translates into the following algebraic statement: Suppose that $I \subset A$ is an ideal generated by a Koszul-regular sequence $f_1, \ldots , f_ r$ of $A$. Then $A \to A/I$ is a perfect ring map. Since $A \to A/I$ is surjective this is a presentation of $A/I$ by a polynomial algebra over $A$. Hence it suffices to see that $A/I$ is pseudo-coherent as an $A$-module and has finite tor dimension. By definition of a Koszul sequence the Koszul complex $K(A, f_1, \ldots , f_ r)$ is a finite free resolution of $A/I$. Hence $A/I$ is a perfect complex of $A$-modules and we win. $\square$

Proof. We can factor $f$ as the composition

where the first morphism is the map $i = (1, f)$ and the second morphism is the projection. Since $Y \to S$ is flat, see Morphisms, Lemma 29.34.9, we see that $X \times _ S Y \to Y$ is perfect by Lemma 37.61.3. As $Y \to S$ is smooth, also $X \times _ S Y \to X$ is smooth, see Morphisms, Lemma 29.34.5. Hence $i$ is a section of a smooth morphism, therefore $i$ is a regular immersion, see Divisors, Lemma 31.22.8. This implies that $i$ is perfect, see Lemma 37.61.7. We conclude that $f$ is perfect because the composition of perfect morphisms is perfect, see Lemma 37.61.4. $\square$

Proof. We will use the criterion of Descent, Lemma 35.22.4 to prove this. By Definition 37.61.2 being perfect is Zariski local on the base. By Lemma 37.61.3 being perfect is preserved under flat base change. The final hypothesis (3) of Descent, Lemma 35.22.4 translates into the following algebra statement: Let $A \to B$ be a faithfully flat ring map. Let $C = A[x_1, \ldots , x_ n]/I$ be an $A$-algebra. If $C \otimes _ A B$ is perfect as an $B[x_1, \ldots , x_ n]$-module, then $C$ is perfect as a $A[x_1, \ldots , x_ n]$-module. This is More on Algebra, Lemma 15.74.13. $\square$

Proof. The problem is local on $X$ and $S$. Hence we may assume that $X = \mathop{\mathrm{Spec}}(B)$, $S = \mathop{\mathrm{Spec}}(A)$ and $f$ corresponds to a pseudo-coherent ring map $A \to B$.

If (1) holds, then $B$ has finite tor dimension $d$ as $A$-module. Then $B_\mathfrak q$ has tor dimension $d$ as an $A_\mathfrak p$-module for all primes $\mathfrak q \subset B$ with $\mathfrak p = A \cap \mathfrak q$, see More on Algebra, Lemma 15.66.15. Then $\mathcal{O}_ X$ has tor dimension $d$ as a sheaf of $f^{-1}\mathcal{O}_ S$-modules by Cohomology, Lemma 20.48.5. Thus (1) implies (2).

By Cohomology, Lemma 20.48.5 (2) implies (3).

Assume (3). We cannot use More on Algebra, Lemma 15.66.15 to conclude as we are not given that the tor dimension of $B_\mathfrak q$ over $A_\mathfrak p$ is bounded independent of $\mathfrak q$. Choose a presentation $A[x_1, \ldots , x_ n] \to B$. Then $B$ is pseudo-coherent as a $A[x_1, \ldots , x_ n]$-module. Let $\mathfrak q \subset A[x_1, \ldots , x_ n]$ be a prime ideal lying over $\mathfrak p \subset A$. Then either $B_\mathfrak q$ is zero or by assumption it has finite tor dimension as an $A_\mathfrak p$-module. Since the fibres of $A \to A[x_1, \ldots , x_ n]$ have finite global dimension, we can apply More on Algebra, Lemma 15.77.5 to $A_\mathfrak p \to A[x_1, \ldots , x_ n]_\mathfrak q$ to see that $B_\mathfrak q$ is a perfect $A[x_1, \ldots , x_ n]_\mathfrak q$-module. Hence $B$ is a perfect $A[x_1, \ldots , x_ n]$-module by More on Algebra, Lemma 15.77.3. Thus $A \to B$ is a perfect ring map by definition. $\square$

Proof. This is more or less immediate from the definition. Namely, let $U = \mathop{\mathrm{Spec}}(A)$ be an affine open of $X$. Then $i^{-1}(U) = \mathop{\mathrm{Spec}}(A/I)$ for some ideal $I \subset A$ and $A/I$ has a finite resolution by finite projective $A$-modules by More on Algebra, Lemma 15.82.2. Hence $i_*\mathcal{O}_ Z|_ U$ can be represented by a finite length complex of finite locally free $\mathcal{O}_ U$-modules. This is what we had to show, see Cohomology, Section 20.49. $\square$

Proof. We claim that Derived Categories of Schemes, Lemma 36.27.1 applies. Conditions (1) and (2) are immediate. Condition (3) is local on $X$. Thus we may assume $X$ and $S$ affine and $E$ represented by a strictly perfect complex of $\mathcal{O}_ X$-modules. Thus it suffices to show that $\mathcal{O}_ X$ has finite tor dimension as a sheaf of $f^{-1}\mathcal{O}_ S$-modules. This is equivalent to being perfect by Lemma 37.61.11. $\square$

Proof. Let $\{ g_ i : X_ i \to X\} _{i \in I}$ be an fppf covering of schemes and let $f : X \to S$ be a morphism such that each $f \circ g_ i$ is perfect. By Lemma 37.60.13 we conclude that $f$ is pseudo-coherent. Hence by Lemma 37.61.11 it suffices to check that $\mathcal{O}_{X, x}$ is an $\mathcal{O}_{S, f(x)}$-module of finite tor dimension for all $x \in X$. Pick $i \in I$ and $x_ i \in X_ i$ mapping to $x$. Then we see that $\mathcal{O}_{X_ i, x_ i}$ has finite tor dimension over $\mathcal{O}_{S, f(x)}$ and that $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ i, x_ i}$ is faithfully flat. The desired conclusion follows from More on Algebra, Lemma 15.66.17. $\square$

Proof. Since $X$ (and hence $Y$) is locally Noetherian all 4 types of regular immersions agree, and moreover we may check whether a morphism is a regular immersion on the level of local rings, see Divisors, Lemma 31.20.8. Thus the result follows from Divided Power Algebra, Lemma 23.7.5. $\square$